# If you have a 1/3 chance of winning one round and you play 7 rounds.

1. Oct 24, 2012

### physicswinner

If you have a 1/3 chance of winning one round and you play 7 rounds. What are the odds you win 4/7 rounds? me and my friends are having a hard time with this question and also i would like to know how you solved it.
Also i'd like to know if a school had 400 kids and there is 365 days in a year. what are the odds of having a persons birthday on each day.
Thanks!

2. Oct 24, 2012

### rcgldr

You'd have to calculate the odds for one of the possible permutations, such as winning on the first 4 rounds and losing on the last 3, then multiply this by the number of permutations of 7 things taken 4 at a time. You could grind out all the probabilities for all the scenarios, 0 wins to 7 wins, in which case the sum of all those probabilities should exactly equal 1.

The second problem with birthdays is similar, but the numbers involved are much larger. One permuation is the first person has a birthday on any day (365/365), the second on a different day than the first (364/365), the third different than the first two (363/365), ... There may be a mathematical trick to simplify the calculations.

Last edited: Oct 24, 2012
3. Oct 25, 2012

### awkward

The second problem is a version of the "Coupon Collector's Problem". The problem is usually stated in terms of the number of samples required to get a complete set of coupons. In your case, the coupons correspond to the days of the year and the kids are the samples. Here is the statement of the solution, in terms of coupons.

N = the number of different types of coupons
T = the number of coupons that need to be collected to obtain a complete set

$$\Pr(T > n) = \sum_{i=1}^{N-1} \binom{N}{i} \left( \frac{N-i}{N} \right) ^n (-1)^{i+1}$$

Source: A First Course in Probability, Seventh Ediiton, by Sheldon Ross, Section 4.1, Example 1e

I think the previous poster confused your question with the Birthday Problem, which is different (what is the probability that two kids have the same birthday?).